Question

An electron that has a velocity with x component 2.6 × 106 m/s and y component 2.4 × 106 m/s moves through a uniform magnetic field with x component 0.029 T and y component -0.14 T. (a) Find the magnitude of the magnetic force on the electron. (b) Repeat your calculation for a proton having the same velocity.

Answer 1

Answer with Explanation:

We are given that

$v_x=2.6\times 10^6 m/s$

$v_y=2.4\times 10^6 m/s$

$B_x=0.029 T$

$B_y=-0.14 T$

a.We have to find the magnitude of the magnetic force on the electron.

$v\times B=\begin{vmatrix}i&j&k\\2.6\times 10^6&2.4\times 10^6&0\\0.029&-0.14&0\end{vmatrix}$

$v\times B=(-0.364-0.0696)\times 10^6 k=-0.4336\times 10^6 k$

Charge on an electron,q=$-1.6\times 10^{-19} C$

$F=q(v\times B)=\mid -1.6\times 10^{-19}(-0.4336)\times 10^6\mid =6.9\times 10^{-14} N$

Force act along positive z- direction.

b.Charge on proton=$q=1.6\times 10^{-19} C$

$F=\mid 1.6\times 10^{-19}(-0.4336)\times 10^6\mid =6.9\times 10^{-14} N$